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University of Warwick 2008/09

Department of Chemistry CH162 Workshop









CH162 Intro to Physical Chemistry

Chemical Kinetics: Workshop

Question 2:



When benzophenone is illuminated with ultraviolet light, it is excited into a singlet

state. Triethylamine acts as a quencher for the singlet state. In a time-resolved laser

spectroscopy experiment, the emission lifetime of this excited state varied with amine

concentration as shown below:



Experiment [Q] (mol dm-3) τ (s)

I 1x10-3 2.5x10-5

II 5x10-3 1.7x10-5

III 1x10-2 1.2x10-5



Determine, either graphically or by inspection, both the quenching rate constant and

fluorescence lifetime.









1

University of Warwick 2008/09

Department of Chemistry CH162 Workshop







Model answer



Question 2:



Note: You are expected to know how to derive the Stern-Volmer equation by using

the steady state approximation to obtain expressions for both fluorescence quantum

yields with/without quencher and then manipulating these equations. See lecture notes

for details. This question tests you ability to establish which equation to use to tackle

this problem and how to use it.



The Stern-Volmer equation relating to lifetimes should be used for this problem. The

Stern-Volmer equation is given by:



1 1

  k Q [Q] (of a similar form as y = c + mx, where c is the intercept (1/T0)

T T0

and m is the gradient (kQ))



where T is the measured emission lifetime of the excited state, T0 is the emission

lifetime in the absence of quencher, kQ is the quenching rate constant and [Q] is the

concentration of quencher.





Graphical solution:



By plotting 1/T versus [Q] both the quenching rate constant (5.0x106 dm3 mol-1 s-1)

and fluorescence lifetime (28.6 μs) may be inferred. The quenching rate constant is

the gradient and the fluorescence lifetime is 1/intercept.





[Q]/moldm^-3 T/s 1/T / s^-1

0.001 2.50E-05 4.00E+04

0.005 1.70E-05 5.88E+04

0.01 1.20E-05 8.33E+04





1.0E+05





y = 5E+06x + 35021

R2 = 0.9999

7.5E+04

1/T /(s^-1)









5.0E+04









2.5E+04









0.0E+00

-0.002 0 0.002 0.004 0.006 0.008 0.01 0.012



[Q]/moldm^-3









2

University of Warwick 2008/09

Department of Chemistry CH162 Workshop







Solution by inspection:



As this is a linear plot with all points lying on the line, we can calculate the quenching

rate constant and fluorescence lifetime.



y 2  y1

The gradient of a straight line is given by:

x 2  x1



From the table;



[Q]/moldm^-3 T/s 1/T / s^-1

0.001 2.50E-05 4.00E+04

0.005 1.70E-05 5.88E+04

0.01 1.20E-05 8.33E+04



we can substitute for both yn and xn as follows:



y2  y1 8.33  104  4x104

  4.81x106  5.0x106 dm3mol-1s-1 obtained graphically

x 2  x1 0.01  0.001



We can now rearrange the following formula to make T0 the subject:



1 1

  k Q [Q]

T T0

1 1

  k Q [Q]

T0 T

1 1

T0  

1   1 

T  k Q [Q]  5

 4.81x106 [0.01]

  1.20x10 

6

 28.4x10 s  28.4s  28.6s obtained graphically



Please remember that units in both this and question 1 are very important. You will be

penalized for absence or incorrect units.









3



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